An object of mass m is dropped from height h above a planet of mass M and radius R.Find an expression for the object's speed as
1 answer:
Answer:
Explanation:
Mass of object = m
Height above planet = h
Mass of planet = M
Radius of planet = R
As we have to find out velocity, so let's apply the law of conservation of energies on initial( when the object was at height) and final( when object hit the surface points.
Initial energy = Final energy
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Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
But the force F
But the electric field intensity due to a point charge Q at a distance r meters away is given by
<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>
Answer:
The force will increase in proportion to the mass of the objects
Explanation:
The acceleration due to gravity is always the same. It is expressed in meters per second squared or m/s². The figure of 9.81 m/s² is an average value that was taken after calculating the acceleration under different surfaces. In fact, the acceleration differs depending on the shape of the part of the earth in relation to the earth's magnetic field and force.
Thus, if one washer was 20 kg, the acceleration being 9.81 m/s² the weight will be:
F = ma
=
If there are there washers, the weight will be:
F = 3 * 20 * 9.81
= 588.6 N